Problem: N different natural numbers are given. If you select four different ones and set them as $ A $, $ B $, $ C $, $ D $, the following formula

$ \ Frac {A + B} {C --D} $



I want to find the maximum value of.




Given N different natural numbers, choose 4 different from them and create a program to find the maximum value of the above formula.



Input

The input is given in the following format.


N
a1 a2 ... aN


The number N (4 ≤ N ≤ 1000) of natural numbers is given in the first line. The value ai (1 ≤ ai ≤ 108) of each natural number is given in the second line. However, the same natural number does not appear more than once (ai ≠ aj for i ≠ j).

Output

Outputs the maximum value of the above formula as a real number for a given N natural numbers. However, the error must not exceed plus or minus 10-5.

Examples

Input

10
1 2 3 4 5 6 7 8 9 10


Output

19.00000


Input

5
22 100 42 3 86


Output

9.78947


Input

6
15 21 36 10 34 5


Output

18.00000


Input

4
100000 99999 8 1


Output

28571.285714

Solution (in CPP):

Solution:
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<utility>
#include<vector>
using namespace std;
double maxfirst(vector<double>,int);
double minfirst(vector<double>,int);
double zeti(double);
int main() {
	int n;
	vector<double>a(1001);
	cin >> n;
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	int mi = 100000001,at;
	for (int i = 0; i < n; ++i) {
		for (int j = i; j < n; ++j) {
			if (a[j] < mi) {
				mi = a[j];
				at = j;
			}
		}
		swap(a[i], a[at]);
		mi = 100000001;
	}
	printf("%.5lf\n", max(maxfirst(a, n), minfirst(a, n)));
	return 0;
}
double maxfirst(vector<double>a,int n) {
	int ma = -1, mi = 100000001, atma, atmi;
	double ans;
	for (int i = 0; i < n - 1; ++i) {
		if (a[i] + a[i + 1] > ma) {
			ma = a[i] + a[i + 1];
			atma = i;
		}
	}
	for (int i = 0; i < n-1; ++i) {
		if (i == atma || i + 1 == atma)continue;
		if (zeti(a[i] - a[i + 1])<mi) {
			mi = zeti(a[i] - a[i + 1]);
			atmi = i;
		}
	}
	ans = (a[atma] + a[atma + 1]) / (zeti(a[atmi] - a[atmi + 1]));
	return ans;
}
double minfirst(vector<double>a,int n) {
	int ma = -1, mi = 100000001, atma, atmi;
	double ans;
	for (int i = 0; i < n - 1; ++i) {
		if (zeti(a[i] - a[i + 1])<mi) {
			mi = zeti(a[i] - a[i + 1]);
			atmi = i;
		}
	}
	for (int i = 0; i < n - 1; ++i) {
		if (i == atmi || i + 1 == atmi)continue;
		if (a[i] + a[i + 1] > ma) {
			ma = a[i] + a[i + 1];
			atma = i;
		}
	}

	ans = (a[atma] + a[atma + 1]) / (zeti(a[atmi] - a[atmi + 1]));
	return ans;
}
double zeti(double a) {
	if (a < 0)a *= -1;
	return a;
}